#### Find the equations of the line passing through the point (3, 0, 1) and parallel to the planes x + 2y = 0 and 3y - z = 0.

Given, a line passes through a point P (3, 0, 1) and is parallel to the planes x + 2y = 0 and 3y - z = 0.

We must find the equation of this line.

Let the position vector of point P be

$\vec{a}=3\hat{i}+0\hat{j}+\hat{k}$

Or,

$\vec{a}=3\hat{i}+\hat{k}.....(i)$

Let us consider the normal to the given planes, that is, perpendicular to   the normal of the plane x + 2y = 0 and 3y - z = 0

Normal to the plane x + 2y = 0 can be given as $\vec{n_{1}}=\hat{i}+2\hat{j}$

Normal to the plane 3y - z = 0 can be given as $\vec{n_{2}}=3\hat{j}-\hat{k}$

So, $\vec{n}$   is perpendicular to both these normals.

So,

$\vec{n}=\vec{n_{1}}\times \vec{n_{2}}$

$\Rightarrow \vec{n}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 0 \\ 0 & 3 & -1 \end{vmatrix}$

Taking the 1st row and the 1st column, we multiply the 1st element of the row $\left (a_{11} \right )$ with the difference of products of the opposite elements $\left (a_{22}\times a_{33}-a_{23} \times a_{32} \right )$, excluding 1st row and 1st column

$\begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21}& a_{22} & a_{23} \\ a_{31}& a_{32} & a_{33} \end{vmatrix}=a_{11}\left ( a_{22} \times a_{33}-a_{23} \times a_{32} \right )$

Here,

$\begin{vmatrix} \hat{i}& \hat{j} & \hat{j} \\ 1& 2 & 0 \\ 0& 3 & -1 \end{vmatrix}=\hat{i}\left ( \left ( 2 \times -1 \right )-\left (0 \times 3 \right ) \right )$

Now, we take the 2nd column and 1st row, and multiply the 2nd element of the row (a??) with the difference of the product of opposite elements $\left (a_{21}\times a_{33}-a_{23} \times a_{31} \right )$

$\begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21}& a_{22} & a_{23} \\ a_{31}& a_{32} & a_{33} \end{vmatrix}=a_{11}\left ( a_{22} \times a_{33}-a_{23} \times a_{32} \right )-a_{12}\left ( a_{21} \times a_{33}-a_{23} \times a_{31} \right )$

Here

$\begin{vmatrix} \hat{i}& \hat{j} & \hat{j} \\ 1& 2 & 0 \\ 0& 3 & -1 \end{vmatrix}=\hat{i}\left ( \left ( 2 \times -1 \right )-\left (0 \times 3 \right ) \right )-\hat{j}\left (\left ( 1 \times -1 \right )-\left ( 0 \times 0 \right ) \right )$

Finally, taking the 1st row and 3rd column , we multiply the 3rd element of the row (a??) with the difference of the product of opposite elements  $\left (a_{22}\times a_{33}-a_{23} \times a_{32} \right )$  excluding the 1st row and 3rd column.

$\begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21}& a_{22} & a_{23} \\ a_{31}& a_{32} & a_{33} \end{vmatrix}=a_{11}\left ( a_{22} \times a_{33}-a_{23} \times a_{32} \right )-a_{12}\left ( a_{21} \times a_{33}-a_{23} \times a_{31} \right )+a_{13}\left ( a_{21} \times a_{32}-a_{22} \times a_{31} \right )$

Here

$\begin{vmatrix} \hat{i}& \hat{j} & \hat{j} \\ 1& 2 & 0 \\ 0& 3 & -1 \end{vmatrix}=\hat{i}\left ( \left ( 2 \times -1 \right )-\left (0 \times 3 \right ) \right )-\hat{j}\left (\left ( 1 \times -1 \right )-\left ( 0 \times 0 \right ) \right )+\hat{k}\left ( \left ( 1 \times 3 \right )-\left ( 2 \times 0 \right ) \right )$

Futher simplifying it,

$\Rightarrow \begin{vmatrix} \hat{i}& \hat{j} & \hat{j} \\ 1& 2 & 0 \\ 0& 3 & -1 \end{vmatrix}=\hat{i}\left ( -2-0 \right )-\hat{j}\left ( -1-0 \right )+\hat{k}\left ( 3-0 \right )\\ \\ \\ \Rightarrow \begin{vmatrix} \hat{i}& \hat{j} & \hat{j} \\ 1& 2 & 0 \\ 0& 3 & -1 \end{vmatrix}=-2\hat{i}+\hat{j}+3\hat{k}\\ \\ \\ \rightarrow \vec{n}=-2\hat{i}+\hat{j}+3\hat{k}$

Therefore, the direction ratio is (-2, 1, 3) …(iii)

We know, vector equation of any line passing through a point and parallel to a vector is $\vec{r}=\vec{a}+\lambda \vec{b}$ where $\lambda \epsilon \mathbb{R}$

Hence, from (i) and (ii),

$\vec{a}=3\hat{i}+\hat{k}\\ \vec{n}=-2\hat{i}+\hat{j}+3\hat{k}$

Putting these vectors in the equation $\hat{r}=\hat{a}+\lambda \hat{n}\\$

We get

$\hat{r}=\left ( 3\vec{i}+\vec{k} \right )+\lambda \left ( -2\hat{i}+\hat{j}+3\hat{k} \right )$

But we know,

$\hat{r}=x\vec{i}+y\hat{j}+z\vec{k}$

Substituting this,

$\left (x\vec{i}+y\hat{j}+z\vec{k} \right )=\left ( 3\hat{i}+\hat{k} \right )+\lambda\left ( -2\hat{i}+\hat{j}+3\hat{k} \right )\\ \\ \Rightarrow \left (x\vec{i}+y\hat{j}+z\vec{k} \right )-\left ( 3\hat{i}+\hat{k} \right )=\lambda\left ( -2\hat{i}+\hat{j}+3\hat{k} \right )\\ \\ \Rightarrow x\hat{i}+y\hat{j}+z\hat{k}-3\hat{i}-\hat{k}=\lambda \left ( -2\hat{i}+\hat{j}+3\hat{k} \right )\\ \\ \Rightarrow \left ( x-3 \right )\hat{i}+y\hat{j}+\left ( z-1 \right )\hat{k}=\lambda \left ( -2\hat{i}+\hat{j}+3\hat{k} \right )\\ \\$

Thus, the required equation of the line is $\left ( x-3 \right )\hat{i}+y\hat{j}+\left ( z-1 \right )\hat{k}=\lambda \left ( -2\hat{i}+\hat{j}+3\hat{k} \right )\\ \\$